Proposition 4. If m≥1 , λ N_m , and ρ_mλ , then S^λ occurs in W_m .

Fix m , abbreviate G=G_m , and let ε be the one-dimensional sign representation, also denoting its character by ε . By Young's rule, conjugation of dominance, and the sign-twist identity in, it is enough to show that W_m⊗ε contains every constituent of M^ ρ_m . We will find the cyclic module generated by v_ R_m ⊗v_ R_m inside an isomorphic image of W_m⊗ε , then map that module onto M^ ρ_m . The row subgroup H_R=∏_ i=1 ^m S_ R_ m,i acts on v_ R_m by ε|_ H_R . Consequently the unnormalized signed diagonal average is exactly ∑_ h∈H_R ε(h)\,h w_m =v_ R_m ⊗u, u=∑_ h∈H_R h v_ C_m . To see that u≠0 , let a_0(i,j)=i and consider the corresponding word e_ a_0 . It is fixed by H_R . Its coefficient in v_ C_m is 1 : within each increasingly ordered column it is the identity assignment 1,2,…,m+1-j . The coefficient of e_ a_0 in every h v_ C_m is therefore also 1 , since h^ -1 e_ a_0 =e_ a_0 . Its coefficient in u is |H_R|=∏_ r=1 ^m r! , which is nonzero. We next identify the alternating line in V_R . Frobenius reciprocity, the induction identity for a sign twist, and Lemma 3 ([Young's rule: support and diagonal multiplicity]) give _ H_R (ε|_ H_R ,_ H_R ^ G V_R) , =_G(_ H_R ^ G ε|_ H_R ,V_R); , =_G(M^ ρ_m ⊗ε,V_R); , =_G(M^ ρ_m ,V_R⊗ε)=1. For the last equality, V_R⊗ε≅S^ ρ_m by self-conjugacy of the staircase, and S^ ρ_m has multiplicity one in M^ ρ_m . As v_ R_m is nonzero and alternating under H_R , it spans the line in. Twisting the diagonal tensor product can be done in its second factor: the reassociation (V_R⊗V_C)⊗ε ⟶V_R⊗(V_C⊗ε), (x⊗y)⊗z x⊗(y⊗z) is a G -isomorphism. Choose a G -isomorphism φ:V_C⊗ε⟶V_R , which exists by and ρ_m^t=ρ_m , and choose 0≠z∈ε . Since u is H_R -fixed, u⊗z is alternating under H_R . It follows from that φ(u⊗z)=c\,v_ R_m for some c≠0 . Tensor with z and apply the reassociation in , followed by ⊗φ . The right-hand side becomes c(v_ R_m ⊗v_ R_m ) . Thus the image of W_m⊗ε under this isomorphism contains Z_m=[G](v_ R_m ⊗v_ R_m ) ⊆V_R⊗V_R. We construct a surjection from Z_m onto M^ ρ_m . Let E_ out have basis f_1,…,f_ 2m-1 and define the linear map q:E_m⊗E_m⟶E_ out , q(e_a⊗e_b)=f_ a+b-1 . Regroup the two input layers by position, using ordinary tensor reassociation, and apply q at every position. Then project onto the span M_α of words in which letter r occurs exactly r times for 1≤r≤m , and no letter greater than m occurs. Thus the content is α=(1,2,…,m,0,…,0) in the output alphabet. Denote the resulting linear map by Q:E_m^ ⊗B_m ⊗E_m^ ⊗B_m ⟶ M_α. Both operations commute with position permutations: the same local map q is used everywhere, and content is unchanged by G . Hence Q is G -equivariant. Consider any summand in the expansion of v_ R_m ⊗v_ R_m . On a row of length r , write its two assignments as (a_1,…,a_r) and (b_1,…,b_r) , each a permutation of 1,…,r , and put c_j=a_j+b_j-1 . Then ∑_ j=1 ^r c_j=r^2, ∑_ j=1 ^r c_j^2-r^3 =∑_ j=1 ^r(c_j-r)^2≥0. For a term retained by the content projection, the sum of squared output letters over the whole triangle is ∑_ r=1 ^m r^3 . Summing over the rows forces equality on every row. Thus c_j=r at every position in the row of length r . There is only one possible retained output word, namely t_m=_ (i,j)∈B_m f_ m+1-i . It remains to compute its coefficient. On a row of length r , each of the r! assignments a_j=π(j) has the unique compatible assignment b_j=r+1-π(j) . The latter is the permutation ω_rπ , where ω_r(j)=r+1-j , so the product of the two alternating signs is (π)(ω_rπ) =(ω_r)=(-1)^ r2 . The row contributions are independent. In particular, there is no cancellation, and the exact image is Q(v_ R_m ⊗v_ R_m ) =((-1)^ ∑_ r=1 ^m r2 ∏_ r=1 ^m r!)t_m≠0. The stabilizer of t_m is exactly H_R , since its equal-letter sets are the rows of B_m . Its orbit is a basis of M_α , and hence Q(Z_m)=[G]t_m= M_α ≅_ H_R ^ G 1≅M^ ρ_m . This also explains why the increasing content (1,2,…,m) gives the usual module indexed by the decreasing partition ρ_m : reordering the content parts conjugates the standard Young subgroup. By Lemma 3 ([Young's rule: support and diagonal multiplicity]), every S^τ with τρ_m occurs in the quotient in. Semisimplicity and imply that it occurs in W_m⊗ε . If ρ_mλ , reversal of dominance under conjugation gives λ^tρ_m^t=ρ_m . Taking τ=λ^t and twisting back by ε proves that S^λ occurs in W_m .